New ShengShou 9x9

[tx789]

About half way through that I'd start scrambling randomly I think ROFL!

only at 70 moves what was he expecting?

Spoiler

2R D' 3R2 2U 2L2 4F' U' 2L 2U 3B2 D 2L2 2F' 2L' 3F F2 4U' 3B L' 3F 3U F2 3U2 2L2 3R 3B 2U' 2B2 3F' L' 2B 2D2 U2 4R2 3B' R2 3D 2L' D' 4U' 4F' 3L 4F' 2U L' 3B2 4U' 2U' 2B 3F 2L2 2B2 2U2 2B2 2D2 R2 2B2 D2 4U' 2U2 R2 4U' L' 4U B' 3B2 3L 2B2 3F' F' 2D' 4F2 3R' 2R' 2U2 3R2 D2 2B 2D2 3U' U 2B 3R 3D2 4U 2R F2 2L 4R' 3R2 2D' 2L2 3R2 F 3U2 B2 4F' 2D 4F2 R 4U U2 3F 3D 3U' 2U 3R2 U 4R2 2F2 F 3R2 2D2 3U' 2U2 4R' R' 4U 2B2 2L B2 2F 3D 3L2 3R2 3D' U' B' 2B' L D 3F2 2D2 B U2 B2 L2 3B 4U' 2B2 F 4U' 3F' 2L B2 3R2 R2 3F 2U' 3B 2F' F2 R' B2 2D2 2B' 3L' 4R' 2D' L' 2L' 3D2 2L 2B 3D 4R2 2U' 4R 2D' 2U U 3B 3F' 3D2 2L' 4U2 L' 3L 4R' 4F' 3U' 2L' 3L 4U2 3L F2 2U2 U2 3B2 U2 4F2 2L2 2R2 2D' 3R' 2B2 3F2 2F2 2L2 2R2 4F2 D' B 4U 2B' D2 3D 4F L2 3U2 L' 2B 4U2 B' 2D 4F R2 3U2 L2 3L2 3R2 4F2 3F' F2 3L 4R 3R2 4U 2L2 3B' 2F2 2U 2F2 3R' 2D 3F 2F F' D' 3B2 3R' 2F' 2D' 2U2 2L2 4R R 4F' 2D' 3D2

still 140 moves is long. I think god number for it may be 80 moves since if you only use R,L,U,D ect it's just like a 3x3 same with the 2R, 2L,2D ect 3L, 3R you get the idea.

 

[LNZ]

You can emulate a 3x3x3 cube using a 9x9x9 cube. Big cubes are great for this.

 

[tx789]

You can emulate a 3x3x3 cube using a 9x9x9 cube. Big cubes are great for this.

in 4 different ways on a 9x9

 

[rahulkadukar]

in 4 different ways on a 9x9

I see what you did there.

Not Bad

 

[qqwref]

8x8x8s are more useful - you can emulate any LxMxN cuboid for L, M, N all <= 8

 

[jla]

Mike should do this BLD

 

[Kattenvriendin]

8x8x8s are more useful - you can emulate any LxMxN cuboid for L, M, N all <= 8

And also solve it that way. Just had an adjacent edge parity. Solved it like on a 4x4 (hold one unsolved on the front, one on the right): r2 U2 r2 Uw2 r2 u2

Just treat the inner two times 3 layers as two times one and presto

 

[rubixwiz031]

This will never become a WCA event.
Remember, speedstacks only go for 10 mins. I doubt more than a few (Lin, Kevin, Michael, Dan, Feliks, Mats, etc) could solve it in that time consistently.

 

[qqwref]

We're moving away from stackmat timers, so in the future the limit may be more like an hour. Still, extremely long speedsolving events are not particularly fun to organize or judge - especially when you know you'll have a lot of new people who know they can solve it eventually but will take a huge amount of time to do so.

 

[CarlBrannen]

I suspect that in the limit of very large cubes, it eventually becomes advantageous to solve at least the last four edges by two stages, orient then permute. The reason is that the algorithm for orienting does not become appreciably more complicated as the size of the cube increases.

Orienting the last four edges puts each edge piece in its correct slice. This means that the number of possible situations with the last two edges is vastly reduced; for an nxnxn cube there are 2^(n-3) possibilities but they have an extremely easy group structure, just the product of a bunch of groups with two elements each.

 

[qqwref]

I don't really see that as an advantage. After the pieces are "orient"ed, how do you propose to "permute" them efficiently? And wouldn't it be sort of icky to recognize the "orient"ation, especially if you don't leave the four E slice edges for last?

Also, solving the last four eges on the 9x9x9 is the same as solving the last four edges on the 5x5x5 three times... so you might as well just use the same strategy three times (unless you see a way to save moves by doing the same thing on multiple slices at once).

 

[Kattenvriendin]

unless you see a way to save moves by doing the same thing on multiple slices at once.

That IS doable, it's what I did on the 8x8 but you have to plan ahead really well.

 

[qqwref]

Uh, I know it's doable, I'm actually a pretty decent bigcube solver :fp

 

[Kattenvriendin]

Your "see a way" can be interpreted in different ways, obviously.

The facepalm wasn't quite needed you know

 

[CarlBrannen]

"Also, solving the last four eges on the 9x9x9 is the same as solving the last four edges on the 5x5x5 three times... so you might as well just use the same strategy three times (unless you see a way to save moves by doing the same thing on multiple slices at once). "

Yes, that's exactly it. You can do multiple slices at once. This happens because the "edge flip/swap" algorithm
(Ll)' U2 (Ll)' U2 F2 (Ll)' F2 (Rr)' U2 (Ll)2
generalizes. See "edge flip/swap" here:
www.bigcubes.com/5x5x5/lastedges.html

It generalizes in that you can use it to flip/swap more than just one pair at a time. You replace the (Ll) and (Rr) moves with something more complicated according to which pairs you wish to swap and which you want to leave alone. The various slices (i.e. the slices made with L and R moves) commute with one another.

In other words, orienting the edges gives a situation where you need to perform some number of "swap/flips" but these can be done in one algorithm.

I should admit that I've not yet haven't trained myself to do this. But it should be clear that you can involve more than one left slice in the algorithm. And the algorithm has two types of moves. There are "outer" moves that consist only of U2 and F2, and slice moves. But if you take the mirror image of the algorithm (i.e. left to right), then U2 and F2 are left unchanged. So the algorithm to flip/swap the right edges is identical to the above except for the slice moves.

This works for different slices. That is, you can flip/swap the L2 and R3 slices at the same time. But it doesn't work for the same slices simultaneously, i.e. L2 and R2. Maybe there's a way to do L2 and R2 simultaneously, right now I use a different algorithm for them anyway.

 

[bluecloe45]

Looks great, maybe I'll get into big cubes

Psych.

 

[TMOY]

I suspect that in the limit of very large cubes, it eventually becomes advantageous to solve at least the last four edges by two stages, orient then permute. The reason is that the algorithm for orienting does not become appreciably more complicated as the size of the cube increases.

Orienting the last four edges puts each edge piece in its correct slice. This means that the number of possible situations with the last two edges is vastly reduced; for an nxnxn cube there are 2^(n-3) possibilities but they have an extremely easy group structure, just the product of a bunch of groups with two elements each.

I already usa a "orient then permute" system for the las 4 edges of big cubes (I use sandwich). But my notion of orientation is different from yours, I consider a wing oriented if its stickers are either on the correct face or on the opposite one. I don't know if it's faster than intuitive edge solving but at least it helps recogntion.

 

[CarlBrannen]

I already usa a "orient then permute" system for the las 4 edges of big cubes (I use sandwich). But my notion of orientation is different from yours, I consider a wing oriented if its stickers are either on the correct face or on the opposite one. I don't know if it's faster than intuitive edge solving but at least it helps recogntion.

Since a wing only has two possible orientations, I'm assuming that what you do is to make sure the wings are oriented correctly according to the center edges.

So I'm thinking you don't need to orient the center edges, instead you fix the wings to match the centers. If I didn't get this right, please correct me.

 

[TMOY]

Well, no. First I orient the outermost wings according to the outer layers, then I permute them, then I solve the outermost remaining wings the same way, and so on. I solve midges (on odd-layered cubes) only after being done with wings.

 

[LNZ]

Some news is at hand. After some delays making the cardboard boxes for the Shengshou 9x9x9, the cube is available to buy from 15/11/2012.

The site 51morefun.com just shipped my white SS9 only hours ago.